Find the vector equation of the line passing through $(1,2,3)$ and parallel to the planes $\vec r \cdot (\hat i - \hat j + 2\hat k) = 5$ and $\vec r \cdot (3\hat i + \hat j + \hat k) = 6.$

.: Let the direction of the line be $\vec b = {b_1}\hat i + {b_2}\hat j + {b_3}\hat k.$

The equation of the line passing through $(1,2,3)$ having the direction$\vec b$ is
$\vec r = (\hat i + 2\hat j + 3\hat k) + \lambda \vec b = (\hat i + 2\hat j + 3\hat k) + \lambda ({b_1}\hat i + {b_2}\hat j + {b_3}\hat k)$ ...(1)

Since line (1) is parallel to $\vec r \cdot (\hat i - \hat j + 2\hat k) = 5.$

$\therefore$ $({b_2}\hat i + {b_2}\hat j + {b_3}\hat k) \cdot (\hat i - \hat j + 2\hat k) = 0$

' $\Rightarrow$ ${b_1} - {b_2} + 2{b_3} = 0$ ….(2)

Again, since line (1) is parallel to $\vec r \cdot (3\hat i + \hat j + \hat k) = 6$ ,

$\therefore$ $({b_1}\hat i + {b_2}\hat j + {b_3}\hat k) \cdot (2\hat i + \hat j + \hat k) = 0$

$\Rightarrow$ $3{b_1} + {b_2} + {b_3} = 0$ ….(3)

Solving (2) and (3),

we get
$\cfrac{{{b_1}}}{{ - 1 - 2}} = \cfrac{{{b_2}}}{{6 - 1}} = \cfrac{{{b_3}}}{{1 + 3}} \Rightarrow \cfrac{{{b_1}}}{{ - 3}} = \cfrac{{{b_2}}}{5} = \cfrac{{{b_3}}}{4}$

$\therefore$ $< - 3,5,4 >$ are direction ratios of the line.

$\therefore$ The required equation of the line is

$\vec r = \hat i + 2\hat j + 3\hat k + \lambda ( - 3\hat i + 5\hat j + 4\hat k)$